Here's a problem I've found entertaining.

Is it possible to find a subset of 3-dimensional Euclidean space such that its homology groups (integer coefficients) or one of its fundamental groups is not torsion-free?

Context: The analogous question has a negative answer in dimension 2. This is a theorem of Eda's (1998). In dimension 4 and higher, the answer is positive as the real projective plane embeds. If the subset of 3-space has a regular neighbourhood with a smooth boundary, a little 3-manifold theory says the fundamental group and homology groups are torsion-free.

edit: Due to Autumn Kent's comment and the ensuing discussion, torsion in the homology has been ruled out provided the subset of $\mathbb R^3$ is compact and has the homotopy-type of a CW-complex (more precisely, if Cech and singular cohomologies agree).

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    $\begingroup$ There's a continuum in $\mathbb{R}^4$ such that for an arbitrary group $G$, it has a subspace with fundamental group $G$. I had trouble finding the paper (it's pretty recent, maybe 5ish years old?), but my idea would be to read that paper and see if anything they did happens to live in $\mathbb{R}^3$, perhaps after a well-chosen quotient. It'd surprise me if you could get ANYTHING in dimension $4$, but NOTHING in dimension $3$. Maybe someone with better Googling skills can link the paper, at least. $\endgroup$ Feb 9, 2021 at 19:38
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    $\begingroup$ @JohnSamples I bet you're thinking of Ziga Virk's paper for countable groups $G$. This is a nice result but unfortunately it doesn't doesn't provide much for the $\mathbb{R}^3$ situation. I'll go ahead and add here that shape/Cech invariants will also be pretty unhelpful to this problem since there many subsets with trivial shape and uncountable first singular homology. $\endgroup$ Feb 22, 2021 at 21:42
  • $\begingroup$ Yup, that's the one! It's a great paper; maybe not the biggest result in the world, but it's just such a pleasure. $\endgroup$ Feb 22, 2021 at 22:27

6 Answers 6


I don't think that

torsion in the homology has been ruled out

Certainly, torsion in Cech cohomology has been ruled out for a compact subset. The "usual" universal coefficient formula, relating Cech cohomology to $\operatorname{Hom}$ and $\operatorname{Ext}$ of Steenrod homology, is not valid for arbitrary compact subsets of $\Bbb R^3$ (although it is valid for ANRs, possibly non-compact). The "reversed" universal coefficient formula, relating Steenrod homology to $\operatorname{Hom}$ and $\operatorname{Ext}$ of Cech cohomology is valid for compact metric spaces, but it does not help, because $\operatorname{Ext}(\Bbb Z[\frac1p],\Bbb Z)\simeq\Bbb Z_p/\Bbb Z\supset\Bbb Z_{(p)}/\Bbb Z$, which contains $q$-torsion for all primes $q\ne p$. (Here $\Bbb Z_{(p)}$ denotes the localization at the prime $p$, and $\Bbb Z_p$ denotes the $p$-adic integers. The two UCFs can be found in Bredon's Sheaf Theory, 2nd edition, equation (9) on p.292 in Section V.3 and Theorem V.12.8.)

The remark on $\operatorname{Ext}$ can be made into an actual example. The $p$-adic solenoid $\Sigma$ is a subset of $\Bbb R^3$. The zeroth Steenrod homology $H_0(\Sigma)$ is isomorphic by the Alexander duality to $H^2(\Bbb R^3\setminus\Sigma)$. This is a cohomology group of an open $3$-manifold contained in $\Bbb R^3$, yet it is isomorphic to $\Bbb Z\oplus(\Bbb Z_p/\Bbb Z)$ (using the UCF, or the Milnor short exact sequence with $\lim^1$), which contains torsion. Of course, every cocycle representing torsion is "vanishing", i.e. its restriction to each compact submanifold is null-cohomologous within that submanifold.

By similar arguments, $H_i(X)$ (Steenrod homology) contains no torsion for $i>0$ for every compact subset $X$ of $\Bbb R^3$.

It is obvious that "Cech homology" contains no torsion (even for a noncompact subset $X$ of $\Bbb R^3$), because it is the inverse limit of the homology groups of polyhedral neighborhoods of $X$ in $\Bbb R^3$. But I don't think this is to be taken seriously, because "Cech homology" is not a homology theory (it does not satisfy the exact sequence of pair). The homology theory corresponding to Cech cohomology is Steenrod homology (which consists of "Cech homology" plus a $\lim^1$-correction term). Some references for Steenrod homology are Steenrod's original paper in Ann. Math. (1940), Milnor's 1961 preprint (published in http://www.maths.ed.ac.uk/~aar/books/novikov1.pdf), Massey's book Homology and Cohomology Theory. An Approach Based on Alexander-Spanier Cochains, Bredon's book Sheaf Theory (as long as the sheaf is constant and has finitely generated stalks) and the paper

As for torsion in singular $4$-homology of the Barratt-Milnor example, this is really a question about framed surface links in $S^4$ (see the proof of theorem 1.1 in the linked paper).

  • $\begingroup$ Are you saying that there is torsion in $H_4$ of Barratt-Milnor ? $\endgroup$
    – BS.
    Nov 14, 2010 at 10:13
  • $\begingroup$ I'm only saying that one who is interested in whether there is torsion in singular $H_4$ of the $2$-dimensional Barratt-Milnor example (which I think is conceivable) might find it helpful to look at my geometric proof of the Barratt-Milnor original result that its singular $H_3$ is nonzero (and in fact uncountable). Personally I am not interested at all, because I don't know of any single true application of singular homology/homotopy beyond the case where it coincides with Steenrod homology/homotopy. (They coincide on spaces that are homotopy equivalent to ANRs, including CW-complexes.) $\endgroup$ Nov 15, 2010 at 19:38
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    $\begingroup$ Perhaps I should clarify that the Barratt-Milnor example, as I see it, shows precisely that singular homology is pathological on compact metric spaces (because it doesn't feel dimension). I wanted to understand why this is so, but haven't signed up for all pathology. There are other ways in which singular homology is pathological on compact metric spaces: it fails the strong excision axiom and the Alexander duality; it doesn't understand in any way the operation of inverse limit; K(Z,n) doesn't represent singular cohomology. Steenrod homology is free of all these deficiencies. Hope this helps. $\endgroup$ Nov 16, 2010 at 11:10
  • $\begingroup$ Dear Sergey, sorry for taking so long to get back on this. Yes I believe you are correct and I stated too much in ruling out torsion. I'll correct that statement. $\endgroup$ Nov 19, 2010 at 19:43
  • $\begingroup$ Hi Ryan, sorry that I messed up dimensions in the original response. $\endgroup$ Nov 19, 2010 at 20:24

I'll assume that the subset is compact.

Then, if you use Cech cohomology, Alexander duality turns this into a question about the complement, which is a 3-manifold.

So, I answer with another question: Can a (wild) open submanifold of the 3-sphere have torsion in its homology? (My guess is no. But then I'm not RH Bing.)

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    $\begingroup$ Ah, that's nice. The answer to your question is no. If you had torsion in H_1 of an open 3-dimensional submanifold of R^3, you'd have it in a compact 3-dimensional submanifold of R^3. That doesn't happen -- I didn't supply the argument but it it boils down to what's known as Fox's Re-Embedding Theorem (which is an application of Dehn's Lemma), that such a 3-manifold can be re-embedded to be the complement of disjoint embedded handlebodies, another duality application says H_1 is free. So part of my question is answered. The fundamental group question remains open. $\endgroup$ Nov 7, 2009 at 2:04
  • $\begingroup$ Oh, very good. I knew the argument for compact submanifolds, but was spacing on whether or not you can immediately jump to a compact submanifold. Thanks. $\endgroup$ Nov 7, 2009 at 2:09
  • $\begingroup$ If a manifold is aspherical (a $K(G,1)$) then its fundamental group cannot have an element of order $n>0$ because then the manifold would have a covering space which is both a manifold and a $K(\mathbb Z/n,1)$. Thus a noncompact $3$-manifold with torsion in $\pi_1$ would have to have $\pi_2\ne 0$. Does that help? $\endgroup$ Jul 13, 2010 at 18:35
  • $\begingroup$ n>1, I meant of course $\endgroup$ Jul 13, 2010 at 23:34
  • $\begingroup$ Tom, I think the Sphere Theorem (that any 3-manifold containing an essential sphere contains an essential embedded sphere) works for any 3-manifold, so a submanifold of $S^3$ with connected complement would be aspherical and so wouldn't have torsion in $\pi_1$. For the problem at hand, you can of course assume the subset of interest, X say, is connected, but it would seem that the issue is that you're interested in the fundamental group of X, and not the complementary 3-manifold (and you don't have duality at your disposal for $\pi_1$. $\endgroup$ Jul 14, 2010 at 0:23

There is the Barratt-Milnor 1962 example of "anomalous (singular) homology", showing that the rational singular homology of the one point union $X$ of countably many spheres $S^2$ whith radius tending to $0$ is non zero in all dimensions $>1$ (and is in fact uncountable). They use Hurewicz maps and infinite sums of Whitehead products of elements of homotopy groups of spheres, but I don't see if torsion in higher $\pi_i(S^2)$ could give torsion in $H_*(X,Z)$.


I think your subset of R^3 must be pretty ugly to have a fighting chance. If it is a compact subpolyhedron of R^3, then by Alexander duality its k-homology is the same as (2-k)-dimensional cohomology of an open 3-manifold. The only interesting case is k=1 because 0th (co)homology are torsion free, but if the open manifold is homotopy equivalent to a finite complex then by universal coefficients 1st cohomology is torsion free. This rules out all "nice" examples.


Proposition. Suppose that a subset $X\subset R^3$ is semilocally simply connected (SLSC). Then for each $x\in X$, the group $G= \pi_1(X,x)$ is torsion-free.

Proof. I will be using the fact that if $U$ is an open connected subset of $R^3$, then $\pi_1(U)$ is torsion-free, see for instance here.

Let $c: S^1\to X$ be a loop representing an element of order $n$ in $\pi_1(X,x)$. Accordingly, let $c_n: S^1\to X$ be the precomposition of $c$ with the map $z\mapsto z^n$ and $h: D^2\to X$ be an extension of $c_n$ to the unit disk. The image $Y:= h(D^2)$ is compact and locally path-connected. Thus, there is a function $\phi(\delta)$ such that if $d(y_1, y_2)\le \delta$, $y_1, y_2\in Y$, then there is a path of diameter $\le \phi(\delta)$ in $Y$ connecting $y_1, y_2$. Furthermore, since $Y$ is compact and $X$ is assumed to be SLSC, there exists $\epsilon>0$ such that if $\alpha: S^1\to Y$ has image of diameter $\le \epsilon$, then $\alpha$ extends to a continuous map $D^2\to X$.

Now, consider the system of open $\frac{1}{i}$-neighborhoods $U_n$ of $Y$ in $R^3$. Let $r: U_i\to Y$ denote a (likely discontinuous) nearest-point projection (defined via the Axiom of Choice). Since each group $\pi_1(U_i,x)$ is torsion-free, the map $c: S^1\to Y$ extends to a (continuous) map $f_i: D^2\to U_i$.

I will now imitate the standard argument (I think, due to Borsuk), from the proof that each finite-dimensional compact locally-contractible metrizable space is ANR.

Given a triangulation $T$ of $D^2$, I define the map $g_i: T^{(0)}\to Y$ as the composition of the restriction of $f_i$ to the $T^{(0)}$ with the projection $r$. The goal is to show that for large $i$, the map $g_i$ extends to a map $D^2\to X$ which restricts to $c$ on $S^1=\partial D^2$.

First of all, if $\frac{1}{i}\le \delta$, and $T$ is such that the diameters of the images under $f_i$ of the edges of $T$ are $\le\delta$, then for each edge $e=[v,w]$ of $T$, $d(g_i(v), g_i(w))\le 3\delta$. Hence, by the local path connectivity of $Y$, we can extend $g_i$ to $e$ so that $g_i(e)\subset Y$ has diameter $\le \phi(3\delta)$. In the case of boundary edges of the disk $D^2$, we will assume that $g_i|_e=c|_e$. Note that for each 2-simplex $\Delta$ in $T$, the diameter of $g_i(\partial \Delta)$ is $\le 3\phi(3\delta)$. By taking $i$ sufficiently large and taking the triangulation $T$ sufficiently fine, we can assume that $3\phi(3\delta)\le \epsilon$, where $\epsilon$ is defined as above. Hence, for this value of $i$, the map $g_i$ extends to a map $g: D^2\to X$. It follows that $[c]=1\in G=\pi_1(X,x)$ and, hence, $G$ is torsion-free. qed


The answer in this post shows that if $U$ is open and connected, its fundamental group must be torsion-free.

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    $\begingroup$ I put this information in my original question, 10 years ago. $\endgroup$ Jun 18, 2019 at 23:06

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